Q: Crushing-law constants: the constants in Bond’s, Rittinger’s, and Kick’s laws (Kb, Kr, Kk) depend primarily on what factors?

Introduction / Context: Energy–size reduction laws (Kick, Rittinger, Bond) use empirical constants to relate energy input to size change. These constants are not universal; they reflect both the material being crushed and the machine/process conditions. Given Data / Assumptions: Kk (Kick), Kr (Rittinger), Kb (Bond) appear in their respective energy relations. We are deciding what they depend on. Concept / Approach: Material properties (hardness, fracture toughness, structure, moisture) strongly affect breakage work. Machine class (jaw, gyratory, roll) and operating conditions (speed, nip, liner profile) also influence the measured constant during testing. Step-by-Step Solution: Recognize that constants are calibrated from tests on specific materials and machines.Therefore, they depend on both feed characteristics and the crusher type/operating conditions.Select “both (a) & (b).” Verification / Alternative check: Work index data are tabulated by material, and test protocols specify equipment—evidence that constants are contextual. Why Other Options Are Wrong: Material only or machine only is incomplete.“Neither” contradicts empirical calibration practice. Common Pitfalls: Treating constants as universal numbers transferable across materials and crushers without recalibration. Final Answer: both (a) & (b)

Q: Bulk solids handling: which conveyor is most suitable for long-distance transport of cold, non-abrasive, granular or fine materials?

Introduction / Context: Choosing a conveyor depends on distance, material abrasiveness, lump size, and capacity. For extended runs moving cold, non-abrasive bulk solids, one option predominates in industry. Given Data / Assumptions: Material: granular/fines, non-abrasive, ambient temperature. Distance: long (tens to hundreds of metres or more). Concept / Approach: Belt conveyors provide high capacity, low specific power, and gentle handling over long distances with minimal transfer points. Alternatives like screw or apron conveyors are better for short distances or very abrasive/hot service; bucket elevators are vertical or steep-incline devices, not for long horizontal runs. Step-by-Step Solution: Match requirement (long distance, gentle handling) → Belt conveyors.Eliminate others: screws for short runs, apron for heavy/abrasive lumps, bucket for elevation.Select “Belt conveyor.” Verification / Alternative check: Plant layouts for grains, fertilizers, and packaged solids commonly deploy belt conveyors for long corridors and overland conveying. Why Other Options Are Wrong: Bucket: vertical service.Screw: short distances; high wear with long runs.Apron: robust but slow/heavy; suited to hot, abrasive lumps. Common Pitfalls: Underspecifying idler spacing and skirt sealing for fines; ignoring belt speed limits for dusting or degradation. Final Answer: Belt conveyor

Q: Hindered settling calculation (Richardson–Zaki): a suspension of glass beads in ethylene glycol settles at 1.7 mm/s; a single bead settles at 17 mm/s. With index n = 4.5, what is the solids volume fraction φ?

Introduction / Context: Hindered settling in suspensions is slower than the terminal settling of a single particle because surrounding particles impede flow. The Richardson–Zaki correlation relates the hindered settling velocity to solids volume fraction. Given Data / Assumptions: Single-particle terminal velocity, ut = 17 mm/s. Hindered settling velocity, u = 1.7 mm/s. Richardson–Zaki index, n = 4.5 (typical for intermediate Re). Concept / Approach: The correlation is u/ut = (1 − φ)^n. Solve for φ from the measured ratio of velocities. Step-by-Step Solution: Compute velocity ratio: u/ut = 1.7/17 = 0.1.Set 0.1 = (1 − φ)^4.5.Take the 4.5th root: 1 − φ = 0.1^(1/4.5) ≈ 0.599.Therefore, φ ≈ 1 − 0.599 ≈ 0.401 ≈ 0.4. Verification / Alternative check: Back-substitute φ = 0.4 → (1 − φ) = 0.6; (0.6)^4.5 ≈ 0.1, consistent with the given velocities. Why Other Options Are Wrong: 0.1: would give u/ut ≈ (0.9)^4.5 ≈ 0.62, much larger than 0.1.0.6: would give u/ut ≈ (0.4)^4.5 ≈ 0.016, too small.“None” is unnecessary since φ ≈ 0.4 matches an option. Common Pitfalls: Confusing mass and volume fraction; using wrong n for the Reynolds-number range; mixing cm/s and mm/s. Final Answer: 0.4

Question 1

Particle shape factor: what is the sphericity of a right circular cylinder with diameter 1 mm and length 3 mm?

Accepted Answer

Introduction / Context: Sphericity is a dimensionless shape factor used in fluid–particle systems (sedimentation, packed beds, filtration). It helps correct correlations for drag, settling velocity, and pressure drop when the particle is not a perfect sphere. Given Data / Assumptions: Cylinder diameter d = 1 mm, length L = 3 mm. Sphericity, psi = (surface area of a sphere having the same volume as the particle) / (actual surface area of the particle). Ends of the cylinder are flat; include both end and lateral areas. Concept / Approach: Compute particle volume and actual area for the cylinder; then find the sphere having the same volume and compute its surface area. Finally, take the ratio Asphere / Aparticle to obtain the sphericity. Step-by-Step Solution: Volume, Vcyl = (pi/4) * d^2 * L = (pi/4) * 1^2 * 3 mm^3 = 0.75pi mm^3.Equivalent sphere diameter, De such that (pi/6)De^3 = Vcyl → De = (6Vcyl/pi)^(1/3) ≈ 1.651 mm.Surface area of equivalent sphere, As = pi * De^2 ≈ 8.56 mm^2.Actual cylinder area, A = lateral + ends = pidL + 2(pid^2/4) = pi13 + pi(1^2)/2 ≈ 10.996 mm^2.Sphericity, psi = As / A ≈ 8.56 / 11.00 ≈ 0.78. Verification / Alternative check: Typical sphericities: spheres = 1; cylinders with L/d ≈ 3 fall around 0.75–0.8, consistent with the computed value. Why Other Options Are Wrong: 0.9: too high; closer to near-spherical shapes.0.6 or 0.5: too low for L/d = 3; would indicate highly elongated or rough particles. Common Pitfalls: Forgetting to include end areas; mixing diameter and radius; using volume ratio instead of area ratio in the definition. Final Answer: 0.78

Question 2

Jaw crusher mechanics: which size-reduction mechanism dominates in jaw crushers handling brittle ores and rocks?

Accepted Answer

Introduction / Context: Crushers reduce solid size via mechanical forces. Understanding the dominant mechanism helps select equipment and predict product size distribution and wear. Given Data / Assumptions: Jaw crusher with fixed and moving jaws forming a V-shaped chamber. Typical duty: primary crushing of brittle, hard materials. Concept / Approach: Jaw crushers primarily apply compressive force. The moving jaw presses the feed against the fixed jaw; the nip compresses and fractures particles along weaknesses. Secondary mechanisms like abrasion occur, but compression dominates the breakage. Step-by-Step Solution: Observe machine kinematics: oscillating jaw closes the gap.Resulting stress state in particles: predominantly compressive.Therefore, the primary size-reduction mechanism is compression. Verification / Alternative check: Manufacturer literature classifies jaw, gyratory, and roll crushers as compression machines; hammer mills and impactors are impact devices; attrition is typical in mills. Why Other Options Are Wrong: Attrition and cutting dominate in mills or knives, not in jaws.Impact is characteristic of hammer/impact crushers. Common Pitfalls: Assuming that because particles bounce or slide, attrition or impact is dominant; in jaw crushers, these are minor compared to compression. Final Answer: compression

Question 3

Crushing terminology check: identify the wrong statement about gape, angle of nip, crushing efficiency, and reduction ratio.

Accepted Answer

Introduction / Context: Standard terms in comminution carry precise meanings used in design and performance evaluation. Mixing them up leads to wrong selections and misleading efficiency figures. Given Data / Assumptions: Conventional definitions for gape, nip angle, crushing efficiency, and reduction ratio. Comparison asks for the incorrect statement. Concept / Approach: Crushing efficiency (in the Rittinger sense) is defined as surface energy created divided by energy supplied to the machine. Because only a tiny fraction of input energy creates new surface, efficiency values are very small. The other listed definitions are standard and correct. Step-by-Step Solution: Gape: largest open distance in a jaw crusher → correct.Angle of nip: geometric angle where a particle is just gripped by rolls → correct.Reduction ratio: size_feed,max / size_product,max → correct.Efficiency statement given flips the ratio; it claims energy absorbed by the solid divided by new surface energy, which is not the standard definition. Verification / Alternative check: Textbook relations show efficiency = (surface energy created) / (energy input), sometimes referenced to energy absorbed by the solid, which is still in the denominator, not the numerator. Why Other Options Are Wrong: (a), (b), (d) align with standard definitions and are not wrong. Common Pitfalls: Confusing efficiency definitions from crushing vs. grinding; misinterpreting “energy absorbed by solid” vs. “energy input to machine.” Final Answer: Crushing efficiency is the ratio of the energy absorbed by the solid to the surface energy created by crushing.

Question 4

Equipment identification: Bradford breaker is most commonly used for size reduction/cleaning of which feed material?

Accepted Answer

Introduction / Context: Different size-reduction machines target specific materials and duties. The Bradford breaker is a rotating cylindrical screen with internal lifters used primarily in the coal industry for sizing and cleaning. Given Data / Assumptions: Classic Bradford breaker design with rotating drum and internal shelves. Goal: identify the standard industrial application. Concept / Approach: Run-of-mine coal often contains shale, stone, and out-of-seam dilution. As the breaker lifts and drops the feed, friable coal fractures and passes through apertures, while harder, larger rock is carried to the discharge for rejection or further handling. This selective action is suited to coal, not to materials like wheat, talc, or dense ores. Step-by-Step Solution: Recognize machine geometry: rotating, lifting, screening action.Match to an application needing simultaneous size reduction and cleaning.Coal preparation fits best; select “coal.” Verification / Alternative check: Coal preparation texts describe Bradford breakers as common for crushing and deshaling at mine sites. Why Other Options Are Wrong: Talc: requires fine milling (e.g., ball mill), not a breaker.Iron ore: breakage requires crushers; screening/beneficiation uses other equipment.Wheat: handled by mills or rollers in food processing. Common Pitfalls: Confusing Bradford breaker with rotary breakers used for municipal waste; while similar, classic Bradford service is coal. Final Answer: coal

Question 5

Rubber processing: which listed device is most suitable for devulcanising rubber scrap and for preparing rubber dispersions/solutions?

Accepted Answer

Introduction / Context: Devulcanisation breaks sulphur cross-links in vulcanised rubber to reclaim and reuse rubber. Preparing stable water dispersions (latex-like) or solvent-based rubber solutions requires intense shear and specialized machines, not just generic mixers or separators. Given Data / Assumptions: Goal 1: devulcanise scrap rubber (reverse cross-linking). Goal 2: produce water dispersions or rubber solutions. Choices: Banbury-type mixer, propeller agitator, Sharples centrifuge, or none of these. Concept / Approach: Practical devulcanisation uses dedicated devulcanisers (e.g., reclaimers, two-roll cracker mills, pan grinders) and chemical/thermal–mechanical processes with peptisers. Creating dispersions uses colloid mills, high-pressure homogenisers, or dissolvers with strong shear—quite different from the listed devices. Step-by-Step Solution: Banbury mixer: excellent for compounding raw rubber with fillers but not a devulcaniser and not ideal for fine dispersions.Propeller agitator: provides mild mixing; insufficient shear for devulcanisation or stable fine dispersions.Sharples centrifuge: a separator; does not devulcanise or dissolve rubber.Hence, among the options given, the correct choice is “none of these.” Verification / Alternative check: Industrial reclaim lines typically specify cracker mills, refiner mills, and chemical reclaiming; dispersion lines specify colloid/homogeniser equipment—none of which appear among the listed options. Why Other Options Are Wrong: Banbury: mixing/compounding only.Propeller: inadequate shear.Sharples: separation, not reaction or dissolution. Common Pitfalls: Assuming any “mixer” can devulcanise; confusing dispersion formation with mere agitation. Final Answer: none of these

Question 6

Flotation performance metric: if the concentrate grade equals the tailings grade, what is the selectivity index?

Accepted Answer

Introduction / Context: The selectivity index (also called separation factor) is used in flotation and gravity separation to compare the enrichment of valuable mineral in concentrate versus tailings. Given Data / Assumptions: C = grade (mass fraction of valuable) in concentrate. T = grade in tailings. Given C = T. Concept / Approach: One common form is SI = (C * (1 − T)) / (T * (1 − C)). This ratio equals 1 when C = T because numerator and denominator are identical. An SI greater than 1 indicates enrichment; less than 1 indicates poor separation relative to random. Step-by-Step Solution: Start from SI = (C * (1 − T)) / (T * (1 − C)).Set C = T → SI = (C * (1 − C)) / (C * (1 − C)) = 1.Therefore, when concentrate and tailings have identical grade, the process exhibits SI = 1 (no selectivity). Verification / Alternative check: Other equivalent forms of SI reduce to unity for C = T as well, reflecting zero separation power. Why Other Options Are Wrong: 0 or ∞: would imply perfect reverse or perfect separation, not the case.0.5: arbitrary and unsupported by the relation. Common Pitfalls: Confusing selectivity index with enrichment ratio or recovery; mixing units—grades must be fractions, not percentages, inside the formula. Final Answer: 1

Question 7

Crusher selection by Taggart’s rule: choose a jaw crusher rather than a gyratory when (tonnes per hour) divided by gape^2 (cm^2) is less than what value?

Accepted Answer

Introduction / Context: Taggart’s empirical rule provides a quick screen for selecting between jaw and gyratory crushers based on capacity requirement and required gape. It is useful during preliminary plant design. Given Data / Assumptions: Metric form using tonnes per hour and gape in cm. Decision threshold separates jaw preference from gyratory preference. Concept / Approach: Taggart’s rule states that a jaw crusher is preferred when (T/hr) / (gape in cm)^2 is less than approximately 0.0184. Larger values indicate a need for the higher-capacity continuous-duty gyratory design. Step-by-Step Solution: Compute index I = throughput / gape^2 (consistent units).Compare I with the empirical threshold ≈ 0.0184.If I < 0.0184 → choose jaw; otherwise → choose gyratory. Verification / Alternative check: Textbook tables reproduce this threshold in various unit systems; in inch–short ton form a different numerical constant appears but the guideline is the same. Why Other Options Are Wrong: 0.00184: too small; would unduly bias toward gyratories.0.184 or 1.84: too large; inconsistent with the commonly cited metric value. Common Pitfalls: Mixing units (cm vs mm or inches); misreading the square on gape; applying the rule outside preliminary screening. Final Answer: 0.0184

Question 8

Crushing-law constants: the constants in Bond’s, Rittinger’s, and Kick’s laws (Kb, Kr, Kk) depend primarily on what factors?

Accepted Answer

Introduction / Context: Energy–size reduction laws (Kick, Rittinger, Bond) use empirical constants to relate energy input to size change. These constants are not universal; they reflect both the material being crushed and the machine/process conditions. Given Data / Assumptions: Kk (Kick), Kr (Rittinger), Kb (Bond) appear in their respective energy relations. We are deciding what they depend on. Concept / Approach: Material properties (hardness, fracture toughness, structure, moisture) strongly affect breakage work. Machine class (jaw, gyratory, roll) and operating conditions (speed, nip, liner profile) also influence the measured constant during testing. Step-by-Step Solution: Recognize that constants are calibrated from tests on specific materials and machines.Therefore, they depend on both feed characteristics and the crusher type/operating conditions.Select “both (a) & (b).” Verification / Alternative check: Work index data are tabulated by material, and test protocols specify equipment—evidence that constants are contextual. Why Other Options Are Wrong: Material only or machine only is incomplete.“Neither” contradicts empirical calibration practice. Common Pitfalls: Treating constants as universal numbers transferable across materials and crushers without recalibration. Final Answer: both (a) & (b)

Question 9

Bulk solids handling: which conveyor is most suitable for long-distance transport of cold, non-abrasive, granular or fine materials?

Accepted Answer

Introduction / Context: Choosing a conveyor depends on distance, material abrasiveness, lump size, and capacity. For extended runs moving cold, non-abrasive bulk solids, one option predominates in industry. Given Data / Assumptions: Material: granular/fines, non-abrasive, ambient temperature. Distance: long (tens to hundreds of metres or more). Concept / Approach: Belt conveyors provide high capacity, low specific power, and gentle handling over long distances with minimal transfer points. Alternatives like screw or apron conveyors are better for short distances or very abrasive/hot service; bucket elevators are vertical or steep-incline devices, not for long horizontal runs. Step-by-Step Solution: Match requirement (long distance, gentle handling) → Belt conveyors.Eliminate others: screws for short runs, apron for heavy/abrasive lumps, bucket for elevation.Select “Belt conveyor.” Verification / Alternative check: Plant layouts for grains, fertilizers, and packaged solids commonly deploy belt conveyors for long corridors and overland conveying. Why Other Options Are Wrong: Bucket: vertical service.Screw: short distances; high wear with long runs.Apron: robust but slow/heavy; suited to hot, abrasive lumps. Common Pitfalls: Underspecifying idler spacing and skirt sealing for fines; ignoring belt speed limits for dusting or degradation. Final Answer: Belt conveyor

Question 10

Hindered settling calculation (Richardson–Zaki): a suspension of glass beads in ethylene glycol settles at 1.7 mm/s; a single bead settles at 17 mm/s. With index n = 4.5, what is the solids volume fraction φ?

Accepted Answer

Introduction / Context: Hindered settling in suspensions is slower than the terminal settling of a single particle because surrounding particles impede flow. The Richardson–Zaki correlation relates the hindered settling velocity to solids volume fraction. Given Data / Assumptions: Single-particle terminal velocity, ut = 17 mm/s. Hindered settling velocity, u = 1.7 mm/s. Richardson–Zaki index, n = 4.5 (typical for intermediate Re). Concept / Approach: The correlation is u/ut = (1 − φ)^n. Solve for φ from the measured ratio of velocities. Step-by-Step Solution: Compute velocity ratio: u/ut = 1.7/17 = 0.1.Set 0.1 = (1 − φ)^4.5.Take the 4.5th root: 1 − φ = 0.1^(1/4.5) ≈ 0.599.Therefore, φ ≈ 1 − 0.599 ≈ 0.401 ≈ 0.4. Verification / Alternative check: Back-substitute φ = 0.4 → (1 − φ) = 0.6; (0.6)^4.5 ≈ 0.1, consistent with the given velocities. Why Other Options Are Wrong: 0.1: would give u/ut ≈ (0.9)^4.5 ≈ 0.62, much larger than 0.1.0.6: would give u/ut ≈ (0.4)^4.5 ≈ 0.016, too small.“None” is unnecessary since φ ≈ 0.4 matches an option. Common Pitfalls: Confusing mass and volume fraction; using wrong n for the Reynolds-number range; mixing cm/s and mm/s. Final Answer: 0.4

Question 11

Bond’s work index: what are the dimensions/units of the work index used in Bond’s crushing and grinding equation?

Accepted Answer

Introduction / Context: Bond’s equation relates the energy required for size reduction to the 80% passing sizes of feed and product. The proportionality factor is the work index, a material property used in mill selection and power estimation. Given Data / Assumptions: Conventional SI/industrial practice. Focus is on units associated with the work index. Concept / Approach: Bond’s work index, Wi, is defined as the energy per unit mass required to reduce a very large feed to a size such that 80% passes 100 micrometres (in the standard test), expressed in kilowatt-hours per metric ton. Step-by-Step Solution: Recall definition: energy per mass → kWh/ton.Check dimensionality: energy (kWh) divided by mass (ton) → consistent with power–time per mass.Select kWh/ton. Verification / Alternative check: Bond grindability tests report Wi values (e.g., 8–20 kWh/ton) in tabulations for ores. Why Other Options Are Wrong: Dimensionless: false—Wi has energy per mass units.kW/ton: power per mass; not the definition.kWh·m^1/2/ton: a misinterpretation of the size terms in the equation. Common Pitfalls: Confusing the units of Wi with the slope/intercept units from linearized plotting methods. Final Answer: kWh/ton

Question 12

Filtration aids: what is the primary benefit of adding a filter aid to a slurry before or during filtration?

Accepted Answer

Introduction / Context: Filter aids (e.g., diatomite, perlite, cellulose fibers) are added to problematic slurries to improve filtrability when cakes are slimy, compressible, or blinding occurs. They can be precoated onto the medium or dosed into the slurry. Given Data / Assumptions: Conventional pressure or vacuum filtration of aqueous slurries. Filter aid used as precoat or body feed. Concept / Approach: Filter aids create a highly permeable, porous initial layer that supports subsequent cake, increasing permeability and reducing specific cake resistance. The net effect is a higher filtration rate at a given pressure drop. While they can mitigate blinding, claims that they always reduce pressure or always “deplug” are too general. Step-by-Step Solution: Add porous inert solids → composite cake with higher voidage and permeability.From Darcy’s law, increased permeability raises volumetric filtration rate at fixed ΔP and viscosity.Hence the principal benefit is faster filtration. Verification / Alternative check: Precoat filtration practices in beverage and chemical industries document higher rates and longer cycles when using appropriate filter aids. Why Other Options Are Wrong: (a) Reduction in required pressure is not universal; many operations run at fixed ΔP.(c) Deplugging is conditional; not guaranteed for all foulants or operating errors.(d) Cake porosity improves but is not guaranteed “high” for every dense particulate system. Common Pitfalls: Overdosing aids (increasing solids load and cost); choosing chemically incompatible aids that contaminate product. Final Answer: Accelerating the rate of filtration

Question 13

Clay processing: which mixing device is vacuum-operated specifically to de-air (remove entrained air from) clay before shaping?

Accepted Answer

Introduction / Context: In ceramics and brick/tile manufacture, clay must be homogenized and free of entrained air before extrusion or pressing. Air pockets cause cracks, blistering, and dimensional defects during drying and firing. Given Data / Assumptions: Industrial clay preparation prior to shaping. Focus on a device that operates under vacuum to deair. Concept / Approach: A vacuum pug mill blends, kneads, and simultaneously removes entrained air by drawing vacuum in the mixing chamber. This produces plastic, uniform, air-free clay suitable for continuous extrusion. Step-by-Step Solution: Identify equipment with both mixing and vacuum deairing functions → vacuum pug mill.Banbury and muller mixers are intensive mixers but typically lack integrated vacuum deairing for clay bodies.Therefore, select “Pug mill.” Verification / Alternative check: Equipment suppliers for ceramic industry list “vacuum deairing pug mills” as standard just before extrusion. Why Other Options Are Wrong: Banbury: designed for rubber/plastics compounding.Muller: sands/clay mixing without integrated vacuum deairing.None: incorrect because a standard device exists (pug mill). Common Pitfalls: Ignoring vacuum level and throughput matching; insufficient moisture control which still traps air despite deairing. Final Answer: Pug mill

Question 14

Conveyor design parameters: the width and speed of a belt conveyor depend primarily on which properties of the handled material?

Accepted Answer

Introduction / Context: Conveyor selection and sizing are driven by the material’s physical characteristics. To prevent spillage and excessive wear, designers choose belt width and speed based on lump size and mass flow considerations. Given Data / Assumptions: Handling cold, free-flowing bulk solids. Goal: determine which material properties govern width and speed. Concept / Approach: Larger lumps require wider belts and larger idler spacings to maintain a cross-section that contains the stream. Higher bulk density increases mass flow for a given volumetric rate, thereby affecting preferred belt speed to keep power and wear within limits. Step-by-Step Solution: Relate lump size → minimum belt width to avoid edge spillage.Relate bulk density → mass throughput = volumetric rate * density; speed set by capacity and acceptable belt loading.Thus, both properties matter; select “both (a) & (b).” Verification / Alternative check: CEMA and DIN standards use lump size classes and bulk density charts to select width and speed. Why Other Options Are Wrong: Either property alone is insufficient for correct sizing.“Neither” is clearly incorrect as design handbooks rely on these inputs. Common Pitfalls: Neglecting surcharge angle and moisture, which also influence cross-sectional loading. Final Answer: both (a) & (b)

Question 15

Classification term: separation of particles of the same density but different sizes by exploiting their different flow (settling/transport) rates is called what?

Accepted Answer

Introduction / Context: In fluid–solid operations, particles of equal density but different sizes can be separated because their drag and settling behavior differ. Recognizing the correct term helps in selecting appropriate equipment. Given Data / Assumptions: Particles have the same density. Separation leverages different settling/transport velocities in a fluid stream. Concept / Approach: Elutriation uses an upward fluid velocity to selectively carry smaller or lighter particles while larger ones fall or remain. It is distinct from simple screening (sizing), which uses geometric apertures, and from flocculation, a particle–particle aggregation process. Step-by-Step Solution: Match the mechanism (hydrodynamic velocity discrimination) to the term.“Elutriation” specifically means removal of fines by an upward stream, often in classifiers.Choose “elutriation.” Verification / Alternative check: Hydraulic classifiers, elutriators, and air classifiers apply elutriation principles to effect size separations for density-equal solids. Why Other Options Are Wrong: Sorting: generic term; lacks the hydrodynamic feature.Sizing: uses screens; geometric separation.Flocculation: chemical/colloidal aggregation, not a hydraulic size split. Common Pitfalls: Confusing elutriation with pneumatic conveying; conflating density separation (e.g., heavy media) with velocity-based size splits. Final Answer: elutriation

Question 16

Producing fine talc powder from granules: which mill is commonly used to achieve the required fine size?

Accepted Answer

Introduction / Context: Talc is a soft mineral used in fillers and paints, needing fine particle sizes (often below 100 micrometres). Crushers produce coarse to medium sizes; mills are used for fine grinding. Given Data / Assumptions: Starting form: talc granules. Target: fine powder suitable for industrial applications. Concept / Approach: Ball mills provide impact and attrition grinding in a rotating shell partially filled with grinding media. Talc, being soft and flaky, breaks down effectively in such mills to produce a controlled fine size with classification. Step-by-Step Solution: Eliminate primary crushers (jaw, gyratory) and roller crusher that give coarse products.Select a fine grinding device capable of sub-100 μm sizes → ball mill.Therefore, choose “ball mill.” Verification / Alternative check: Industrial talc circuits commonly use ball mills or vertical roller mills with downstream classifiers to meet product specs. Why Other Options Are Wrong: Crushers: designed for coarse reduction, not fine powder. Common Pitfalls: Overgrinding leading to excessive fines and energy use; neglecting closed-circuit classification. Final Answer: ball mill

Question 17

Gyratory vs jaw crushers: which statement is true of gyratory crushers compared to reciprocating jaw crushers?

Accepted Answer

Introduction / Context: Gyratory and jaw crushers are both primary crushers, but their kinematics and continuous vs. intermittent action differ, affecting capacity and power draw. Given Data / Assumptions: Standard large primary machines handling hard rock. Comparison focuses on capacity and operation mode. Concept / Approach: Gyratories crush continuously as the mantle gyrates within the concave, offering higher capacities for a given discharge area. Jaw crushers crush intermittently as the movable jaw oscillates, leading to lower throughput for the same gape and a more pulsating power draw. Step-by-Step Solution: Identify operation mode: gyratory = continuous; jaw = intermittent.Capacity implication: continuous action yields greater capacity per discharge area in gyratories.Therefore, the true statement is option (a). Verification / Alternative check: Manufacturer catalogs show higher tph ratings for gyratories than comparable jaw units at similar open-side settings. Why Other Options Are Wrong: (b) Intermittent crushing describes jaws, not gyratories.(c) Gyratories are well suited for coarse, high-capacity duties.(d) Gyratories have steadier, not less steady, power consumption due to continuous crushing. Common Pitfalls: Judging capacity solely by gape without considering continuous vs. intermittent action and chamber profile. Final Answer: They have greater capacity per unit of discharge area.

Question 18

Short-distance conveying of loose, non-abrasive materials (e.g., garbage, grain, food waste): which conveyor is most suitable?

Accepted Answer

Introduction / Context: For localized transfers inside plants—like under-hopper discharge, short horizontal lifts, or inclined feeds—compact conveyors with simple containment are preferred for loose, non-abrasive materials. Given Data / Assumptions: Materials: garbage, grain, food wastes (non-abrasive, moderate moisture). Distance: short (typically < 30 m per flight). Concept / Approach: Screw conveyors are enclosed, compact, and cost-effective for short runs. They meter and convey simultaneously, tolerate variable feed, and prevent spillage and dusting. Belts excel at long distances; flights/drag conveyors are for heavier, abrasive, or sticky service. Step-by-Step Solution: Match application (short, enclosed, metered) → screw conveyor.Eliminate alternatives based on distance and material abrasiveness.Select “Screw conveyor.” Verification / Alternative check: Facility layouts in grain and food plants commonly use screws under bins, mixers, and for short transfers. Why Other Options Are Wrong: Belt: better for long runs.Flight/drag: robust but higher wear and power for this gentle service. Common Pitfalls: Using screws for very long distances (inefficient) or very abrasive solids (rapid wear). Final Answer: Screw conveyor

Question 19

Filtration plotting: a straight line is obtained by plotting reciprocal filtration rate (dV/dt)^{-1} versus filtrate volume V for which combination of cake compressibility and flow regime?

Accepted Answer

Introduction / Context: Filtration analysis often linearizes the governing equation to estimate specific cake resistance and medium resistance. One common linear plot uses the reciprocal filtration rate as a function of filtrate volume. Given Data / Assumptions: Constant-pressure filtration. Darcy’s law applies (laminar flow through porous media). Cake is incompressible (resistance proportional to thickness). Concept / Approach: For incompressible cakes under laminar flow, the Ruth equation gives t = (μ * α * C / (2 * A^2 * ΔP)) * V^2 + (μ * Rm / (A * ΔP)) * V, where symbols have their usual meanings. Differentiating, the reciprocal filtration rate (dV/dt)^{-1} varies linearly with V with slope proportional to μ * α * C / (A^2 * ΔP) and intercept proportional to μ * Rm / (A * ΔP). Step-by-Step Solution: Start from constant-ΔP filtration with incompressible cake: t = aV^2 + bV.Differentiate: dt/dV = 2aV + b → (dV/dt)^{-1} = 2aV + b, which is linear in V.Therefore, a straight line in the indicated plot implies laminar flow with an incompressible cake. Verification / Alternative check: With compressible cakes, α depends on pressure and cake thickness, introducing nonlinearity; turbulent flow also violates Darcy’s law linearity. Why Other Options Are Wrong: Compressible cake: slope/intercept vary with V and ΔP → nonlinearity.Turbulent conditions: Darcy’s law no longer holds; empirical exponents distort linear plots. Common Pitfalls: Plotting t/V vs V and (dV/dt)^{-1} vs V interchangeably; both give linear forms for incompressible cakes but with different coefficients—ensure the correct plot is used. Final Answer: incompressible cake and laminar flow

Question 20

Size reduction basics — reduction ratio for fine grinders:
In comminution terminology, what is the typical reduction ratio achievable by fine grinders (i.e., equipment used for fine to ultrafine size reduction)?

Accepted Answer

Introduction / Context: Reduction ratio is a quick way to characterise how much a size-reduction device can shrink particle size in one stage. It is defined as characteristic feed size divided by characteristic product size. Coarse crushers achieve modest ratios, while fine grinders (e.g., ball/tube mills with classifiers, jet mills) are designed to push the fineness much further. This question reinforces typical magnitudes so that you can rapidly select equipment during preliminary design. Given Data / Assumptions: “Fine grinders” means mills intended to produce fine to ultrafine products (well below typical crusher product sizes). One-stage reduction performance is being compared qualitatively. Representative characteristic sizes (e.g., 80% passing) are implied. Concept / Approach: Jaw and gyratory crushers often deliver ratios around 3–8. Secondary/tertiary cone or impact stages may reach the teens. Fine grinding devices can achieve much higher ratios because the starting feed is already small and the product is pushed into powder ranges. In standard handbooks, fine grinders are commonly cited with reduction ratios up to about 100 in a single stage, depending on classifier efficiency and circuit configuration. Step-by-Step Solution: Identify the class: fine to ultrafine grinding equipment.Recall typical ranges: crushers 3–8; intermediate 8–20; fine grinders up to ~100.Choose the largest realistic single-stage ratio listed: “As high as 100.” Verification / Alternative check: Vendor data sheets for fine mills and closed-circuit systems frequently show feed in hundreds of micrometres reduced to single-digit micrometres, consistent with ratios approaching two orders of magnitude. Why Other Options Are Wrong: 5–10; 10–20; 20–40: typical of crushers or coarse/intermediate grinding, not fine grinders. > 500 in one pass: unrealistic without multi-stage or extreme classification recycle. Common Pitfalls: Equating multi-stage cumulative reduction with single-stage capability; always check if the ratio refers to a single pass or an overall circuit. Final Answer: As high as 100

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